Difference between revisions of "2000 USAMO Problems/Problem 1"
(→Solution 2) |
(→Solution 2) |
||
| Line 30: | Line 30: | ||
However, by the very convex condition, | However, by the very convex condition, | ||
<cmath>f\left(2^{-n}\right) + f\left(-2^{-n}\right) \ge \frac{1}{2^{n-2}}.</cmath> | <cmath>f\left(2^{-n}\right) + f\left(-2^{-n}\right) \ge \frac{1}{2^{n-2}}.</cmath> | ||
| − | This is a contradiction. It follows that | + | This is a contradiction. It follows that no very convex function exists. |
== See Also == | == See Also == | ||
Revision as of 23:57, 4 April 2016
Contents
Problem
Call a real-valued function 
very convex if
![\[\frac {f(x) + f(y)}{2} \ge f\left(\frac {x + y}{2}\right) + |x - y|\]](http://latex.artofproblemsolving.com/4/e/2/4e298e1ba811b64c1726bdb5ffab3dddb4f284e0.png)
holds for all real numbers 
and 
. Prove that no very convex function exists.
Solution 1
Let 
, and substitute 
. Then a function is very convex if 
, or rearranging,
![\[\left[\frac{f(a+2b)-f(a+b)}{b}\right]-\left[\frac{f(a+b)-f(a)}{b}\right] \ge 4\]](http://latex.artofproblemsolving.com/6/a/3/6a36f41315521af133a2e6d89575640cccf81ad2.png)
Let 
, which is the slope of the secant between 
. Let 
be arbitrarily small; then it follows that 
, ![$g(a+2b) - g(a+b) > 4,\, \cdots, g(a+kb) - g(a+ [k-1]b) > 4$](http://latex.artofproblemsolving.com/1/8/f/18f2a7263fc48c992600c2b9731235f69438fb3c.png)
. Summing these inequalities yields 
. As 
(but 
, so 
is still arbitrarily small), we have 
. This implies that in the vicinity of any 
, the function becomes vertical, which contradicts the definition of a function. Hence no very convex function exists.
Solution 2
Suppose, for the sake of contradiction, that there exists a very convex function 
Notice that 
is convex if and only if 
is convex, where 
is a constant. Thus, we may set 
for convenience.
Suppose that 
and 
By the very convex condition,
![\[\frac{f(0) + f\left(2^{-n}\right)}{2} \ge f\left(2^{-(n+1)}\right) + \frac{1}{2^n}.\]](http://latex.artofproblemsolving.com/4/f/e/4fe81548506e6fc4eca7a40ed2026ba37bb46016.png)
A straightforward induction shows that:
![\[f\left(2^{-n}\right) \le \frac{A - 2n}{2^n}\]](http://latex.artofproblemsolving.com/1/e/8/1e85d6678d6c56e85f60417c61e644cb64dabdcf.png)
for all nonnegative integers 
Using a similar line of reasoning as above,
![\[f\left(-2^{-n}\right) \le \frac{B - 2n}{2^n}.\]](http://latex.artofproblemsolving.com/8/6/c/86ca5e851710d7fda339ef1889086bf7b043a0b1.png)
Therefore, for every nonnegative integer 
![\[f\left(2^{-n}\right) + f\left(-2^{-n}\right) \le \frac{A+B-4n}{2^n}.\]](http://latex.artofproblemsolving.com/6/c/d/6cd7a1ed4af4348b784133280fde9d94120917b2.png)
Now, we choose 
large enough such that 
This is always possible because 
and 
are fixed for any particular function It follows that:
![\[f\left(2^{-n}\right) + f\left(-2^{-n}\right) < \frac{1}{2^{n-2}}.\]](http://latex.artofproblemsolving.com/e/c/8/ec8db41701559e800fc945e2dd2314eb9c7e02a7.png)
However, by the very convex condition,
![\[f\left(2^{-n}\right) + f\left(-2^{-n}\right) \ge \frac{1}{2^{n-2}}.\]](http://latex.artofproblemsolving.com/6/6/f/66f88ac610104a2f840cd3a25dcb26ae6761aed3.png)
This is a contradiction. It follows that no very convex function exists.
See Also
| 2000 USAMO (Problems • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
